42 watts seems ok for a very small installation, right? I'm thinking enough to charge a phone and maybe a laptop. And, that would be driveable with a small 12V battery and a small solar panel.
As others have pointed out voltage drop is a significant problem with low voltage systems. #24 is 84 ohms per km, or 0.084 ohms per metre. Say the system is running at 13.4 V and the loads will work on as little as 11.5V before they brown out, that means we can have 13.4-11.5V voltage drop = 1.9V voltage drop. ohms law V=IR so for 3.5A and 1.9V R=0.54 ohms total resistance, and at .084 ohms per metre we can transmit 3.5A 6.46 metres before the voltage drop will be unacceptable.
You could charge a phone, but even my small laptop chargers are over 60W. The big ones are 90W.
Also don't try this in your house! pushing 3.5A through that little pinner wire it could get hot and start a fire.
Voltage drop is one way to look at it, but since efficiency is paramount, it's better to look at losses. Losses in the conductor are I^2 * R, so when there's a current of 3.5A, losses are 6.6W at an input of 3.5*13.4=46.9W. Efficiency is 88% which is very bad. An average inverter that converts the low DC voltage to AC is much more efficient and even with taking into account the wiring loss, there is a net gain. Generally, using a low voltage DC grid isn't a good idea.
But i forgot about the resistance from the load back to the source so above where I said 6m should actually be 3m. Or if two wires are parallel both from source to load and load back to source then we're back to 6m.
